Shirts are 5% off. The original price of one shirt is $15. What is the total cost, in dollars, of a shirt at the sale price, inc
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Answer:
98.75% probability that every passenger who shows up can take the flight
Step-by-step explanation:
For each passenger who show up, there are only two possible outcomes. Either they can take the flight, or they do not. The probability of a passenger taking the flight is independent from other passenger. So the binomial probability distribution is used to solve this question.
However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
The probability that a passenger does not show up is 0.10:
This means that the probability of showing up is 1-0.1 = 0.9. So
Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets
This means that
Using the approximation:
(a) What is the probability that every passenger who shows up can take the flight
This is , so this is the pvalue of Z when X = 120.
has a pvalue of 0.9875
98.75% probability that every passenger who shows up can take the flight